Answer:
Given function: [tex]f(x)=(x-2)^2 (x+1) (x-4)[/tex]
The factors give us the zeros (x-intercepts) of the graph.
To find the zeros, set the function to zero and solve for x:
[tex](x-2)^2 (x+1) (x-4)=0[/tex]
Therefore,
[tex](x-2)^2=0\implies x-2=0 \implies x=2[/tex]
[tex](x+1)=0\implies x=-1[/tex]
[tex](x-4)=0 \implies x=4[/tex]
So the zeros are at (2, 0) (-1, 0) and (4, 0)
The multiplicity of a zero is the number of times the factor appears in the fully factored form of the polynomial, i.e. the exponent on the corresponding factor.
- [tex]x = 2[/tex] has a multiplicity of 2
- [tex]x=-1[/tex] has a multiplicity of 1
- [tex]x=4[/tex] has a multiplicity of 1
If the zero has an even multiplicity, the graph touches and bounces off the x-axis at that point.
If the zero has an odd multiplicity, the graph crosses the x-axis at that point.
To find the y-intercept, expand the function:
[tex]f(x)=x^4-7x^3+12x^2+4x-16[/tex]
Set [tex]x = 0[/tex] and solve:
[tex]f(0)=(0)^4-7(0)^3+12(0)^2+4(0)-16=-16[/tex]
Therefore, the y-intercept is at (0, -16)
The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity.
As the leading degree is 4 (even) and the leading coefficient is positive, the end behavior of the function is:
[tex]f(x) \rightarrow + \infty \textsf{ as } x \rightarrow - \infty[/tex]
[tex]f(x) \rightarrow + \infty \textsf{ as } x \rightarrow + \infty[/tex]
